Problem: Balance the following chemical equation: $ $ $\text{CaCl}_2 +$ $\text{Na}_3\text{PO}_4 \rightarrow$ $\text{Ca}_3\text{(PO}_4\text{)}_2 +$ $\text{NaCl}$
Answer: We can treat the phosphate polyatomic ion $\text{(PO}_4\text{)}$ as an atom, symbolized by ${X}$ $ \text{CaCl}_2 + \text{Na}_3{\text{PO}_4} \rightarrow \text{Ca}_3({\text{PO}_4\text})_2 + \text{NaCl} $ $ \text{CaCl}_2 + \text{Na}_3{X} \rightarrow \text{Ca}_3{X}_2 + \text{NaCl} $ There is $1 \space X$ on the left and $2 \space X$ on the right, so multiply $\text{Na}_3X$ by ${2}$ $ \text{CaCl}_2 + {2}\text{Na}_3X \rightarrow \text{Ca}_3X_2 + \text{NaCl} $ There are $6 \text{ Na}$ on the left and only $1$ on the right, so multiply $\text{NaCl}$ by ${6}$ $ \text{CaCl}_2 + 2\text{Na}_3X \rightarrow \text{Ca}_3X_2 + {6}\text{NaCl} $ That gives us $6 \text{ Cl}$ on the right and only $2$ on the left, so multiply $\text{CaCl}_2$ by ${3}$ $ {3}\text{CaCl}_2 + 2\text{Na}_3X \rightarrow \text{Ca}_3X_2 + 6\text{NaCl} $ Now $\text{Ca}$ is balanced too. Replacing $\text{PO}_4$ for $X$, the balanced equation is: $ 3\text{CaCl}_2 + 2\text{Na}_3\text{PO}_4 \rightarrow \text{Ca}_3\text{(PO}_4\text{)}_2 + 6\text{NaCl} $